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# Forces Acting on an Object

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Three forces acting on an object are given by F1 = (âˆ’1.85 i + 7.35 j) N, F2 = (4.70 i âˆ’ 1.15 j) N, and F3 = (âˆ’45.5 i) N. The object experiences an acceleration of magnitude 3.70 m/s2.
(a) What is the direction of the acceleration?
(degrees counterclockwise from the positive x axis)
(b) What is the mass of the object in kg?
(c) If the object is initially at rest, what is its speed after 16.0 s in m/s?
(d) What are the velocity components of the object after 16.0 s in m/s?

https://brainmass.com/physics/newtons-laws/forces-acting-object-547720

## SOLUTION This solution is FREE courtesy of BrainMass!

F1 = (-1.85 i + 7.35 j) N,
F2 = (4.70 i - 1.15 j) N,
F3 = (-45.5 i) N.

Net force,
Fnet = F1 + F2 + F3
= (-1.85 i + 7.35 j) + (4.70 i - 1.15 j) + (-45.5 i)
= (-1.85 + 4.70 - 45.5)i + (7.35 - 1.15)j
= (-42.65 i + 6.20 j) N == (Fx i + Fy j)

(a) Direction of force and acceleration will be the same = ?
Let us assume, angle with +x axis = theta (counter clock wise)
tan(theta) = Fy/Fx = 6.20/(-42.65) = -0.145
=> theta = arctan(-0.145) = 180 - 8.27 = 171.73 degree

(b) mass (m) = ?
|F| = m*|a|
|a| = 3.7 m/s^2
|F| = sqrt(42.65^2 + 6.2^2) = 43.1 N

Hence,
m = |F|/|a| = 43.1/3.7 = 11.6 kg

(c)
Initial velocity, u = 0
acceleration, a = 3.7 m/s^2
time, t = 16.0 s
velocity, v = ?
v = u +at
v = 0 + 3.7*16 = 59.2 m/s

(d)
x-component of velocity,
v_x = v*cos(theta) = 59.2*cos(171.73) = -58.58 m/s

y component of velocity,
v_y = c*sin(theta) = 59.2*sin(171.73) = 8.52 m/s

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!